pontifexmaximus
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The question gives a merry-go-round with a radius of 1m and mass 75kg
The moment of inertia = 0.5(75)(1)^2 = 37.5
A person pushes the merry-go-round with a force of 300N therefore 300 = 37.5 x angular acceleration
therefore angular acceleration = 8 rad s^-2

However if a child weighing 20kg sits 0.75 m away from the centre, what is the angular acceleration?

I am looking for help with working this out not just the answer, if anyone could help that would be great.
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GgbroTG
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I'll admit that I'm not entirely certain of this approach, but we'll try it anyway!

So, modelling the child as a point particle (pffft), its moment of inertia about its centre of mass is 0. However, it's moment of inertia about the centre of the merry-go-round is given by the parallel-axis theorem:



I_c = I_c^\prime + Md^2

\\\

I_c^\prime = 0 \therefore 

\\\

I_c = Md^2

\\\

I_c = 20 \cdot 0.75^2

\\\

I_c = 11.25 \ {\rm kgm}^2

Therefore, the total moment of inertia about the centre of the merry-go-round is:



I_T = I_m + I_c

\\\

I_T = 48.75 {\rm kgm}^2

Then, assuming that the same force is applied at the rim of the merry-go-round,



\overrightarrow{\tau} = \vec{r} \times \vec{F}

\\\

\overrightarrow{\tau} = I \vec{\alpha}

Since the direction isn't really important (I assume), I'll drop the vector notation for my own sake:



rF = I \alpha

\\\

\alpha = \frac{rF}{I}

\\\

\alpha = \frac{(1)300}{48.75}

\\\

\alpha = 6.15 \ {\rm rad \ s}^{-2}

Or, to 1 significant figure (assuming radius is 1m and not 1.00m or anything):



\alpha = 6 \ {\rm rad \ s}^{-2}
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Eimmanuel
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(Original post by GgbroTG)
I'll admit that I'm not entirely certain of this approach, but we'll try it anyway!

So, modelling the child as a point particle (pffft), its moment of inertia about its centre of mass is 0. However, it's moment of inertia about the centre of the merry-go-round is given by the parallel-axis theorem:

 

I_c = I_c^\prime + Md^2 

\\\ 

I_c^\prime = 0 \therefore  

\\\ 

I_c = Md^2 

\\\ 

I_c = 20 \cdot 0.75^2 

\\\ 

I_c = 11.25 \ {\rm kgm}^2

Therefore, the total moment of inertia about the centre of the merry-go-round is:

 

I_T = I_m + I_c 

\\\ 

I_T = 48.75 {\rm kgm}^2

Then, assuming that the same force is applied at the rim of the merry-go-round,

 

\overrightarrow{\tau} = \vec{r} \times \vec{F} 

\\\ 

\overrightarrow{\tau} = I \vec{\alpha}

Since the direction isn't really important (I assume), I'll drop the vector notation for my own sake:

 

rF = I \alpha 

\\\ 

\alpha = \frac{rF}{I} 

\\\ 

\alpha = \frac{(1)300}{48.75} 

\\\ 

\alpha = 6.15 \ {\rm rad \ s}^{-2}

Or, to 1 significant figure (assuming radius is 1m and not 1.00m or anything):

 

\alpha = 6 \ {\rm rad \ s}^{-2}

Since the direction isn't really important (I assume), I'll drop the vector notation for my own sake:
Directions are important as far as vector cross product operation is concerned. Is the torque slowing down or speeding up the angular speed? The cross product can tell us.
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GgbroTG
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(Original post by Eimmanuel)
Directions are important as far as vector cross product operation is concerned. Is the torque slowing down or speeding up the angular speed? The cross product can tell us.
While true, I was referring to the answer itself - as to whether the direction should be included (given that it wasn't included in their answer to the first part of the question).
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Eimmanuel
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(Original post by GgbroTG)
While true, I was referring to the answer itself - as to whether the direction should be included (given that it wasn't included in their answer to the first part of the question).
Even the question does not state the direction, it does not really mean the students do not need to state their choice of direction.
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GgbroTG
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(Original post by Eimmanuel)
Even the question does not state the direction, it does not really mean the students do not need to state their choice of direction.
Absolutely fair. The only reason I chose to 'ignore' the direction was that I was getting tired of writing vector notation in LaTeX. I suppose a more correct way of avoiding vector notation would be to state the components explicitly,



\tau_z = r_x F_y
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Eimmanuel
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(Original post by Leah.J)
Can either of you help me understand Pascal’s principle please ?
the definition says that if a pressure is applied on a liquid it is transmitted undiminished through the liquid but I don’t understand what that means ?
does the internal pressure of the fluid itself increase ? Does Pascal’s principle just say that the pressure of every particle in the fluid will increase by the same amount ?

also, in this picture
Attachment 866382
Attachment 866380
the article says that the equation only works if the height of the pistons is the same, why ? To me it makes perfect sense that in the diagram I drew ( red and green) although the pistons aren’t at the same height, F1/A1=F2/A2
why not ? Help please, I know it’s simple and I don’t know why I’m struggling
Please don't piggyback on other people thread to ask your question. It is very rude and not respecting OP. Start your own thread and you can tag the user to seek for help.
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Leah.J
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(Original post by Eimmanuel)
Please don't piggyback on other people thread to ask your question. It is very rude and not respecting OP. Start your own thread and you can tag the user to seek for help.
I deleted it
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